How to prove that $\lfloor{\sqrt{n}\rfloor} = \Theta(\sqrt{n})$?

Asked Sep 23 '16 at 22:10

Active Sep 23 '16 at 22:58

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It is quite direct to prove that $\lfloor{\sqrt{n}\rfloor} = O(\sqrt{n})$. But how can I prove that $\lfloor{\sqrt{n}\rfloor} = \Omega(\sqrt{n})$?

edited Sep 23 '16 at 22:58

Mario Cervera

asked Sep 23 '16 at 22:10

Roshan SA

0 Answers0